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Work hardening question 2

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A Squared

Aerospace
May 10, 2021
2
Hi all, I've got a question about work hardening I'm hoping someone can shed some light on. The context is driving aluminum aircraft rivets. (2117-t4 Aluminum alloy) They are set cold with a rivet gun, whcih is essentially a pneumatic hammer. I have read numerous sources whcih tell you that you don't want to use too "light" a rivet gun for the size rivet, because a lighter gun will work harden the rivet before its fully upset. A "light" gun, delivers lighter hammer blows, more rapidly vs a heavier gun whcih delivers harder hammer blows at a lower cyclic rate. So, I'm wondering, is this accurate, that a lighter gun will work harden the rivet more than a heavier gun? It seems to me that work hardening would be proportional to the amount of plastic deformation, not how hard or how rapidly you hit it. In other words, if you took a 1/8" diameter, 1/4" long aluminum cylinder and upset it to 3/16" diameter and 1/8" long, you'd get the same amount of work hardening regardless of whether you used 30 light hammer blows, or 10 much harder hammer blows. Is work hardening more complex than just a matter of how much you deform the metal?
 
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There are a long list of factors at play in this.
Yes metals care what increment the work is and not just total strain. There are examples of metals that work both ways on this.
Aloys also respond differently depending on the strain rate, that is how quickly the deformation happens. Some are very sensitive to this and some not.
And there is internal heating cause by the work and in Al rivets this starts to trigger the aging process.
So in this this case fewer but larger impacts ends up delivering the best results.

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P.E. Metallurgy, consulting work welcomed
 
Thanks EdStainless. I really appreciate the explanation.
 
The same concern applies the machining work hardening metals, where it is important to use relatively large cut depth per pass to get through the work hardened layer left by the previous pass.
 
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